ISI2021-MMA: 5

Question

If $(^{n}C_{0} + ^{n}C_{1}) (^{n}C_{1} + ^{n}C_{2}) \cdots (^{n}C_{n-1} + ^{n}C_{n}) = k \;  ^{n}C_{0} \; ^{n}C_{1} \cdots \; ^{n}C_{n-1},$ then $k$ is equal to

• A. $\frac{(n+1)^{n}}{n!}$
• B. $\frac{n^{n}}{n!}$
• C. $\frac{(n+1)^{n}}{nn!}$
• D. $\frac{(n+1)^{n+1}}{n!}$

Comments

Put $n=3$, we get k = $16*2/3$. It satisfies option A.

Answer 1

Answer : (A)

First let’s simplify ${n \choose k} + {n \choose k+1} = \frac{n!}{(n-k)! * k!} + \frac{n!}{(n-k-1)!*(k+1)!}$

                                                         $= \frac{(k+1)n! + (n-k)n!}{(n-k)!*(k+1)!} = \frac{n!(k+1+n-k)}{(n-k)!*(k+1)!}$

                            $\therefore {n\choose k} + {n \choose k+1} = \frac{(n+1)*n!}{(n-k)!*(k+1)!}$    ------------  (eq 1)

 

Now simplifying LHS using (eq 1)

$({n\choose 0} + {n\choose 1})({n\choose 1} + {n\choose 2}) … ({n\choose n-1} + {n\choose n}) = (\frac{(n+1)*n!}{n!*1!})(\frac{(n+1)*n!}{(n-1)!*2!})...(\frac{(n+1)*n!}{(1)!*n!})$   ------------ (eq 2)

 

Solving RHS

$k*{n\choose 0}*{n \choose 1}*...*{n\choose n-1} = k*(\frac{n!}{n!*0!})*(\frac{n!}{(n-1)!*1!})*(\frac{n!}{(n-2)!*2!})*...*(\frac{n!}{1!*(n-1)!})$  ------------  (eq 3)

 

Equating (eq 2) and (eq 3)  we get –

$(\frac{(n+1)*n!}{n!*1!})(\frac{(n+1)*n!}{(n-1)!*2!})...(\frac{(n+1)*n!}{(1)!*n!})  = k*(\frac{n!}{n!*0!})*(\frac{n!}{(n-1)!*1!})*(\frac{n!}{(n-2)!*2!})*...*(\frac{n!}{1!*(n-1)!})$

Cancelling terms from LHS and RHS we get

$\frac{(n+1)^n}{n!} = k*1$

$\therefore k = \frac{(n+1)^n}{n!}$